Problem: Kana was just an ordinary high school girl before a talent scout discovered her. Then, she became an idol. But different from the stereotype, she is also a gameholic. 

One day Kana gets interested in a new adventure game called Dragon Quest. In this game, her quest is to beat a dragon.

<image>

The dragon has a hit point of x initially. When its hit point goes to 0 or under 0, it will be defeated. In order to defeat the dragon, Kana can cast the two following types of spells. 

  * Void Absorption

Assume that the dragon's current hit point is h, after casting this spell its hit point will become \left⌊ h/2 \right⌋ + 10. Here \left⌊ h/2 \right⌋ denotes h divided by two, rounded down.

  * Lightning Strike

This spell will decrease the dragon's hit point by 10. Assume that the dragon's current hit point is h, after casting this spell its hit point will be lowered to h-10.




Due to some reasons Kana can only cast no more than n Void Absorptions and m Lightning Strikes. She can cast the spells in any order and doesn't have to cast all the spells. Kana isn't good at math, so you are going to help her to find out whether it is possible to defeat the dragon.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next t lines describe test cases. For each test case the only line contains three integers x, n, m (1≤ x ≤ 10^5, 0≤ n,m≤30) — the dragon's intitial hit point, the maximum number of Void Absorptions and Lightning Strikes Kana can cast respectively.

Output

If it is possible to defeat the dragon, print "YES" (without quotes). Otherwise, print "NO" (without quotes).

You can print each letter in any case (upper or lower).

Example

Input


7
100 3 4
189 3 4
64 2 3
63 2 3
30 27 7
10 9 1
69117 21 2


Output


YES
NO
NO
YES
YES
YES
YES

Note

One possible casting sequence of the first test case is shown below:

  * Void Absorption \left⌊ 100/2 \right⌋ + 10=60.
  * Lightning Strike 60-10=50.
  * Void Absorption \left⌊ 50/2 \right⌋ + 10=35.
  * Void Absorption \left⌊ 35/2 \right⌋ + 10=27.
  * Lightning Strike 27-10=17.
  * Lightning Strike 17-10=7.
  * Lightning Strike 7-10=-3.

Solution (in PYTHON3):

Solution:
## necessary imports
import sys
import random
from math import log2, log, ceil
input = sys.stdin.readline

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if a > 0:
        return gcd(b%a, a)
    return b

## prime factorization
def primefs(n):
    ## if n == 1
    if n == 1:
        return 1
    ## calculating primes
    primes = {}
    while(n%2 == 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function
def find(x, link):
    while(x != link[x]):
        x = link[x]
    return x

# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, size, link):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)] 
spf_sieve() 
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().split()))

################# ---------------- TEMPLATE ENDS HERE ---------------- #################

for _ in range(int(input())):
    x, n, m = int_array();
    if x <= 10*m:
        print('YES');
        continue;
    while(n > 0):
        x = x//2; x += 10;
        n -= 1;
    if x <= 10*m:
        print('YES'); 
    else:
        print('NO')